# Bodies connected by the cord and pulley – application of Newton’s law of motion problems and solutions

1. Two boxes are connected by a cord running over a pulley. Ignore the mass of the cord and pulley and any friction in the pulley. Mass of the box 1 = 2 kg, mass of the box 2 = 3 kg, acceleration due to gravity = 10 m/s2. Find (a) The acceleration of the system (b) The tension in the cord! Solution Known :

Mass of the box 1 (m1) = 2 kg

Mass of the box 2 (m2) = 3 kg

Acceleration due to gravity (g) = 10 m/s2

Weight of the box 1 (w1) = m1 g = (2)(10) = 20 Newton

Weight of the box 2 (w2) = m2 g = (3)(10) = 30 Newton

Solution :

(a) magnitude and direction of the acceleration

w2 > w1 so the box 2 accelerates downward and the box 1 accelerates upward.

Forces that has the same direction with acceleration (w2 and T1), its sign positive. Forces that has opposite direction with acceleration (T2 and w1), its sign negative.

F = m a

w2 – T2 + T1 – w1 = (m1 + m2) a ——-> T1 = T2 = T

w2 – T + T – w1 = (m1 + m2) a

w2 – w1 = (m1 + m2) a

30 – 20 = (2 + 3) a

10 = 5 a

a = 10 / 5

a = 2 m/s2

Magnitude of the acceleration is 2 m/s2.

(b) The tension force

The box 2 :

There are two forces acts on the box 2 : first, weight of the box 2 (w2), points downward so it’s positive. Second, tension force exerted on the box 2 (T2), points upward so it’s negative. Apply Newton’s second law of motion.

F = m a

w2 – T2 = m2 a

30 – T2 = (3)(2)

30 – T2 = 6

T2 = 30 – 6

T2 = 24 Newton

Box 1 :

There are two forces acts on the box 1. First, weight of the box 1 (w1), points downward so it’s negative. Second, the tension force exerted on the box 1 (T1) points upward so it’s positive. Apply Newton’s second law of motion :

F = m a

T1 – w1 = m1 a

T1 – 20 = (2)(2)

T1 – 20 = 4

T1 = 20 + 4

T1 = 24 Newton

Magnitude of the tension force = T1 = T2 = T = 24 Newton

[irp]

2. An object on a rough horizontal surface. Mass of the object 1 = 2 kg, mass of the object 2 = 4 kg, acceleration due to gravity = 10 m/s2, coefficient of the static friction = 0.4, coefficient of the kinetic friction = 0.3. The system is at rest or accelerated ? If the system is accelerated, find the magnitude and direction of the system’s acceleration! Solution Known :

Mass of the object 1 (m1) = 2 kg

Mass of the object 2 (m2) = 4 kg

Acceleration due to gravity (g) = 10 m/s2

Coefficient of the static friction (μs) = 0.4

The coefficient of the kinetic friction (μk) = 0.3

Weight of the object 1 (w1) = m1 g = (2)(10) = 20 Newton

Weight of the object 2 (w2) = m2 g = (4)(10) = 40 Newton

Normal force exerted on the object 1 (N) = w1 = 20 Newton

Force of the static friction exerted on the object 1 (fs) = μs N = (0.4)(20) = 8 Newton

Force of the kinetic friction exerted on the object 1 (fk) = μk N = (0.3)(20) = 6 Newton

Wanted: acceleration (a)

Solution :

w2 > fs (40 Newton > 8 Newton) so the object 2 is accelerated vertically downward and the object 1 is accelerated horizontally rightward. The friction force that acts on the objects 1 is the force of the kinetic friction (fk). Apply Newton’s second law of motion :

F = m a

w2 – the = (m1 + m2) a

40 – 6 = (2 + 4) a

34 = 6 a

a = 34 / 6 = 17 / 3

a = 5.7 m/s2

Magnitude of the acceleration = 5.7 m/s2

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